Obviously there has a path between $v_1$ and $v_n$ in $G_P$ whose length is $n-1$.I guess for any $P\in S$,there must exist another path between $v_1$ and $v_n$ in $G_P$ whose length is larger than $n-1$,is it true？

If you suppose that P is (or contains) a perfect matching, then this should follow from Smith's theorem which claims that in any cubic graph G there must be an even number of Hamiltonian cycles through any fixed edge uv. Probably the general version can also be reduced to this case.
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domotorpDec 18 '13 at 7:05

Yes，I know that and I have tried hard to do what you said.But I can not work it out and I believe if my guess is right,there must exist some different method for this problem.
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user40096Dec 18 '13 at 8:28

2

If I understand correctly, you are just considering all graphs on $n$ vertices with minimum degree 1, then subdividing each edge and adding a path through all original vertices?
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nvcleempDec 18 '13 at 8:35

I am sorry,nvcleemp.I do not quite understand what you mean.Can you explain your meaning in detail?
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user40096Dec 19 '13 at 0:52